计算不定积分∫xarctanxdx

2024-11-22 03:14:40
推荐回答(2个)
回答(1):

∫xarctanxdx=x²/2arctanx-1/2x+1/2arctanx+c。c为积分常数。

解答过程如下:

∫xarctanxdx

=∫arctanxdx²/2

=x²/2arctanx-∫x²/2darctanx

=x²/2arctanx-1/2∫x²/(1+x²)dx

=x²/2arctanx-1/2∫(x²+1-1)/(1+x²)dx

=x²/2arctanx-1/2∫1-1/(1+x²)dx

=x²/2arctanx-1/2x+1/2arctanx+c

扩展资料:

分部积分:

(uv)'=u'v+uv'

得:u'v=(uv)'-uv'

两边积分得:∫ u'v dx=∫ (uv)' dx - ∫ uv' dx

即:∫ u'v dx = uv - ∫ uv' d,这就是分部积分公式

也可简写为:∫ v du = uv - ∫ u dv

常用积分公式:

1)∫0dx=c 

2)∫x^udx=(x^(u+1))/(u+1)+c

3)∫1/xdx=ln|x|+c

4)∫a^xdx=(a^x)/lna+c

5)∫e^xdx=e^x+c

6)∫sinxdx=-cosx+c

7)∫cosxdx=sinx+c

8)∫1/(cosx)^2dx=tanx+c

9)∫1/(sinx)^2dx=-cotx+c

10)∫1/√(1-x^2) dx=arcsinx+c

回答(2):

∫xarctanxdx
=1/2∫arctanxd(x²)
=x²/2·arctanx-1/2∫x²/(1+x²)dx
=x²/2·arctanx-1/2∫[1-1/(1+x²)]dx
=x²/2·arctanx-x/2+1/2·arctanx+C
=(x²+1)/2·arctanx-x/2++C